CareerCup

Number of a's = (int)k/26;
Terminating Charter = k%26, where
1 = a
2 = b
.
.
.
26 = z

So, if k=28
No. of a's = (int)28/26 = 1
Terminating Character = 28%26 = 2
Hence result = ab

Can be solved use a recursive method:


import java.util.HashMap;

public class FindTheKthString {

private static HashMap<Integer, StringBuffer> letterCode = new HashMap<Integer, StringBuffer>();

static {
for (int i = 1; i < 26; i++) {
letterCode.put(i, new StringBuffer(Character.toString((char) (i + 96))));
letterCode.put(0, new StringBuffer(Character.toString((char) (26 + 96))));
}
}


private static StringBuffer getTheStringAtKthPosition(int i) {
int lastIndex = i%26;
int step = (i-1)/26;

if(step==0)
{
return new StringBuffer(letterCode.get(lastIndex));
}

else
{
return getTheStringAtKthPosition(step).append(letterCode.get(lastIndex));

}

}


public static void main(String[] args) {
for(int index=700;index<=740;index++)
{
System.out.println(index+" "+getTheStringAtKthPosition(index));
}
}
}

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map.Entry;

public class abcdQuestion {

public static void main(String[] args) {
printString(28);

}

public static void printString(int k) {
if (k <= 26) {
System.out.println(Character.toString((char) (96 + k)));

return;
}

HashMap<Integer, String> hp = new HashMap<Integer, String>();
HashMap<Integer, String> temp = new HashMap<Integer, String>();

for (int i = 1; i <= 26; i++) {

hp.put(i, Character.toString((char) (i + 96)));

}
int count = 27;
int l = 1;

while (count <= k) {

String s = hp.get(l);

Iterator<Entry<Integer, String>> it = hp.entrySet().iterator();
while (it.hasNext() && count <= k) {
String d = s + it.next().getValue();

temp.put(count, d);
count++;

}
hp.putAll(temp);

if ((count - 1) == k) {
System.out.println(hp.get(k));
return;
}

l = l + 1;
}

}
}

#include <stdio.h>
#include <stdlib.h>

int main(){

int ar[10];
int n,t,q,r,k,i;
scanf("%d",&n);

k=0;
r = n%26;
ar[k] = r;
q = n/26;
while(q != 0){
k++;
r = q%26;
ar[k] = r;
q = q/26;
}
printf("Binary:\n");
t=k;
while(t >= 0)
printf("%d | ",ar[t--]);

while(k >= 0){
i=k-1;
if(ar[i] != 0){
t = ar[k--] + 96;
printf("%c",t);
}
else{
if(ar[k] > 1){
t = ar[k] + 95;
printf("%c",t);
}
t = 26 + 96;
printf("%c",t);
k=k-2;
}
}

return;
}

Assumption user will enter +ve value starting from one.

void InterviewQuestions::findValueInSequence()
{
    std::string outputStr = "a";
    int positionRequested = 28;
    std::cout<<"\nEnter the position requested (+ve number only starting from one) : ";
    std::cin>>positionRequested;
    // Decrement as value is calculated as per the index 0;
    --positionRequested;
    int i = 0;
    int n = positionRequested/26;
    int rem = positionRequested%26;
    while ( i < n )
    {
        if (outputStr.size() == 1)
        {
            outputStr.insert(0,1,'a');
        }
        else
        {
            char ch = outputStr[0];
            if ( ch == 'z')
            {
                outputStr[0] = 'a';
                outputStr.insert(0,1,'a');
            }
            else
            {
                ch += 1;
                outputStr[0] = ch;
            }
        }
        i++;
    }
    // add the reminder to the first character of the output string.
    char ch = outputStr[outputStr.size() - 1];
    for ( i = 0; i< rem; i++)
    {
        ++ch;
    }
    outputStr[outputStr.size() - 1] = ch;
    std::cout<<"\nRequested Character is : "<<outputStr.c_str()<<"\n";
}

Use Trie DS , do a level order traversal till the count which you need, and then return the value for that node.

Trie can help here.

given value (say n).
n/26=x and n%26=y.
now what U hv to go is to the x level node and yth element. that O(1) complexity

I believe this is the answer they were looking for:

int num = 28;
char letter;
string answer;

while(num)
{
      letter = (num-1)%26;
      answer.push_back(letter + 'a');
      num = (num-letter)/26;
}

You then must print in reverse order due to the way it is pushed onto the string...

def getStringAtKth(k):
if k<=26:
return chr(96+k)

q,r = divmod(k,26) #q-> quotient and r-> remainder

if r == 0:
q, r = q-1, 26

retstr = getStringAtKth(q)+chr(96 + r)
return retstr

It is much like a 2D array

[   a    b   c   d    z]
[  aa ab ac ad    az]
[  ba bb bc bd    bz]
[---------------------]
[---------------------]
[  za zb zc zd    zz]

let XY is the string at kth position
X can be generate with k/26(0-null, 1-a, 2-b........26-z )
Y can be generate with k%26(0-a, 1-b, 2-c........25-z )

----Prefer algo rather then code in this forum

def find_repr(num):
    s=""
    q,r=num,0
    while True:
       r=q%26
       q=q/26
       if r==0:
          r=26
          q=q-1
       s=chr(r-1+ord('a'))+s
       if q==0: break
    return s

if __name__=='__main__':
   for i in range(703,1000):
      print find_repr(i),
   print

public class Base26 {
public static void main(String[] a) {
String alphabet = "zabcdefghijklmnopqrstuvwxy";
String result = new String();
int num = 26*26*26*26;
if(num == 0) {
System.out.print("a");
} else {
while(num > 0) {
int digit = num%26;
result = alphabet.charAt(digit) + result;
if(digit == 0) {
num = (num-1)/26;
} else {
num = num/26;
}
}
System.out.print(result);
}
}
}

seems like a 26 radix problem

#include <iostream>

using namespace std;

int main()
{
	int k = 28;
	int n = 26;

	k--;

	while(k/n > 0)
	{
		k-=n;
		n*=26;
	}
	n/=26;
	while(n>0)
	{
		char c = k/n+'a';
		cout<<c;
		n/=26;
	}
	cout<<endl;
}

public static String findStringAt2(int position)
	{
		
		String strToReturn = "";
		int p1 = position;
		
		while( p1 > 26)
		{
			p1 = p1 - 26;
			strToReturn += "A";
		}
		
		return strToReturn + c1[p1-1];
		
	}

I not sure of the solution. Do a simple conversion as we do in Decimal to binary with base as 26 instead if 2.

eg: If k is 1408, then

26| 1408
--------
26| 54 - 4
--------
26| 2 - 2
---------

So the solution is 224 and converting it into alphabets gives bbd.

For given eg, k=28 => 12 => ab

Pls correct me

#include<stdio.h>

void print_chars(int k);

int main()
{
int k;

printf("\n enter the value of k \n");
scanf("%d",&k);



print_chars(k);


return 0;
}


void print_chars(int k)
{
if(k==0) return;

print_chars(k/26);

printf("%c",(char)('a'+(k%26-1)));
}

convert input number to 1+ 26 + 26^2 +26^3 +.......26^n + k where k<26^n+1

Convert k to base 26.

prepend enough 0s to make sure its length n.

Convert to corresponding characters. 0-->a..... 25 -->z

It's a base 26 conversion Here is the Java solution:

public static void main(String args[]) {
        int position = 85;
        String result = "";
        while (position > 0) {
            int b = position % 26;
            position = position / 26;
            result = b + result;
        }
        String charResult = "";
        for (int iCount = 0; iCount < result.length(); iCount++) {
            charResult = charResult + (char) (result.charAt(iCount) - 47 + 95);
        }
        System.out.println("String is: " + charResult);
    }

void func(char* res_str, k)
{
	for((k / (26*26)) times)
	{
		append (k/26) to res_str;
	}
	append (k%26) to res_str;
}

{
#include<stdio.h>
int main()
{
int n=703;

char arr[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

while(n){
printf("%c ",arr[(n)%26]);
n=(n)/26;

}
return 0;
}

}
guys is this as simple as this or am i missing something .... of course the above program prints the required atring in reverse we can avoid that by using recursion or store it in a string to reverse it ...

public String encode(final int number) {
StringBuilder btr = new StringBuilder();
int quotient = number;
int remainder = 0;
while(quotient != 0) {
remainder = quotient % base;
if(remainder == 0)
remainder = base;
quotient = (quotient - remainder) / base;

char c = (char)('a' + remainder - 1);
btr.append(c);
}
return btr.reverse().toString();
}

This question can be reduced to converting a decimal number (base 10) to a base 26 number.
Below is the code for it :

#include <iostream>

using namespace std;

int getLength(int number, int base)
{
    int length = 0;
    while(number > 0 )
    {
                 number /= base;
                 length++;
    }    
    return length;
}

int* convertDecimal(int toBase, int number)
{
     int len = getLength(number,toBase);
     int *base26 = new int[len];
     int remainder;
     for(int i=0; i < len ; i++)
     {
                     base26[i] = number % toBase;
                     number /= toBase;
     }
     return base26;
}

int main()
{
    int number ;
    cout<<"Enter the number : ";
    cin>>number;
    int *base26 = convertDecimal(26,number-1);
    int length = getLength(number-1,26);
    for(int i=length-1; i>=0; i--)
    {
            cout<<"  "<<char(97+base26[i]);
    }
    cout<<"\n\n";
    system("pause");
}

/**
	 * 
	 * @param num
	 * @return
	 */
	public static String getExcelHeaderForNum(int num) {
		int prevChar = (num - 1) / 26;
		int thisChar = (num - 1) % 26;

		if (prevChar == 0) {
			return "" + (char) (thisChar + 'A');
		} else {
			return getExcelHeaderForNum(prevChar) + (char) (thisChar + 'A');
		}

	}

}

int n;
	cout<<"NUMBER : ";
	cin>>n;
	int repeat=n/27;
	for(int i=0;i<repeat;i++)
		printf("%c",'a'+i);
	int rem=n%27;
	printf("%c",'a'+rem);

What a LOL question